I am very new to LED wiring in my 3D printing projects and need some advice on the one I am now attempting. Is this the correct forum to ask?
If yes, the pic roughly shows what I have and what I am attempting. Three strings of four LED bulbs that I want to wire through a switch and (hopefully) power with a 9V battery.
Thanks again for the helpful tips, that got this project finished. I ended up going with the design I mentioned in a follow-up post....3 strings of 4 LEDs each, wired in parallel, with a 270ohm resistor, powered by two 9-volt batteries.
Here are the pics for my Flux Capacitor on FB: https://www.facebook.com/ukcat/posts/10155547742687197?notif_id=1530370274090677
I VERY MUCH appreciate everyone's feedback so far. At this point in my project, I can wire it anyway I like, but do prefer taking advantage of what I've done so far as originally pictured, all 4 in parallel.
If I am starting to understand this a bit more, I am looking at adding a 2nd 9v battery and clipping them together for 18v of supply. Using that info, this config in the attached new pic is most desirable for me. Does it look correct?
Right! If it makes sense for you, you could add a fifth LED in serial, but that's only true with 2 x 9 V batteries. The switch should then be from + of the battery to the wire of the resistors
the D in LED stands for diode. and a diode has an exponential function regarding current depending on voltage. once you are above the threshold voltage a small increase in voltage. (depending on the led: e.g 0.4v might double the current). thus it is very hard to predict the exact current when you apply a fixed voltage. get a little to high and you will exceed the allowed current (typically 20mA) and damage your led. so you need either ca current source (constant current) but typically you limit the current with a resistor.
alos having LEDs in parallel is not ideal: a small variation in their characteristic will lead to different currents flowing through them and thus different brightness. so an example: lets say you have a 9V battery and want a 3.2V white LED with 15mA. so you put a resistor in front of the LED: 9V-3.2V is 5.8V and to get 15mA you will 5.8V/15mA = 0.387kOhm. or 380Ohm. so you would e.g. choose 390Ohm.
Now small changes in the characteristic of the LED will still result in a different current but with the resistor it is no longer a problem. e.g. if the LED only need 3.1V instead of 3.2V the current would only increase by 0.1/5.8*100% or 1.7%. In the case where you would apply a fixed 3.2V the current might have incerase 300%.
now the downside of large resitors is: you are wasting energy there. in the exmaple above the LED gets 3.2v15mA which is 48mW where your restistor eats 5.8V15mA which is 87mW. A compromise would be to put LEDs in seriers here.
2 LEDs in series would mean 2*3.2V is 6.4V so for the resitor the voltage would be 9V-6.4V = 2.6V. at 15mA this would be: 0.173kOhm or 172Ohm so you would use: 180Ohm there. so you get the same brightness of the LEDs but only wasting 40%
with 3 white LEDs in seriers you would be on 9.6V. so the voltage is below the threshold voltage and the current would be rather small.
if you do not want to waste power and still guerantee a constant current then you need a more complicated curicuit that pulses the LEDs (and stores engery in an inductor).
I would recommend to use the first version from wkarraker with the two LEDs in serial connection. Depending on the quality of the LEDs some of them might be lighter or dimmer if connected in parallel like in the second example picture. The current limiting resistor should always be used. If you would have made your test with a power supply your LEDs would already have died. Your luck was the inner resistance of the battery that limited the current through the LEDs.
There are LED calculators for online and offline use, a site that can calculate the values for you is below. If you have access to a smartphone you can also download an LED calculator that makes calculating a current limiting resistor a breeze. Hope this helps out!
I've run a couple of calculations through the site. Connecting them in parallel may work against you, if you can break the LEDs into groups of 2 LEDs you can extend the runtime of your 9v battery by quite a bit. I've included the diagram the site created below, you can visit the site and try other combinations if you wish.
Not sure why your comment was "flagged for moderation"....Lord knows when they'll get to it....luckily, I got the text in an email.
OK, so I believe I have soldered the stems together in a parallel method....(newbie here)......all of the positives & negatives are soldered together in those rows. They do light up when I put the stems on a 9v battery.
I'm still trying to wrap my old man brain around the resistor calculations. I've visited several websites, but am not 100% sure of the numbers I need to input for this particular design.
Three banks of four lights each....so the forward voltage for each bank would be 12v? I can wire these banks any way I need to at this point.
EDIT: I believe I have found a calculator that I understand. Shows I have to use two 9v batteries (clipped together) and a 240Ohm resistor.....does that sound right?
Yes, you have a parallel circuit for your LEDs. You still need to add a current limiting resistor to the circuit for the LEDs to be protected, connecting all three groups to the battery without is not advisable. I've created a second diagram, trying to modify the groups on the site didn't work out very well so I hacked it with Photoshop. :)
You cite the voltage drop across the LEDs as 3v. This would limit you to running 2 of them from a 9v battery (as a rule of thumb, you don't want the total voltage drop to exceed 2/3 of your supply voltage), unless you include a voltage booster circuit.
Also, speaking generally, you may want to consider a current-limiting resistor (unless the ones you're using have one built in -- most LEDs don't). The formula to determine the value of it in ohms is (Vs - Vf) / I, where Vs is the supply voltage, Vf is the voltage drop (summed if you wire them in series, but your image shows 4 in parallel -- and are those 4 banks wired in series? If so, then you're looking at about 9Vf), and I is the maximum current the LED bank can handle (in amps). If you're wiring all 4 banks, I'd probably include a limiting resister for each bank, and it would be the sum of the max current each LED can handle, which would be 80mA per bank).
For example, for a single 3v, 20ma LED running from a 9V battery, the resistor needed would be (9-3)/0.02, or 300 ohms.